Page 265 - Advanced engineering mathematics

7.10 Linear Transformations 245

Then
⎛ ⎞
1 −1 2 8
0 1 −1 0
⎜ ⎟
⎜ ⎟
1 0 0 −1
⎜ ⎟
⎜ ⎟
A T = 0 1 0 4 ⎟ .
⎜
⎜ ⎟
5 5 −1 0
⎜ ⎟
⎜ ⎟
⎝ 0 0 0 0 ⎠
0 0 0 0
We obtain T (x, y, z,w) as the matrix product
⎛ ⎞
x
y
⎜ ⎟
A T ⎜ ⎟ .
⎝ z ⎠
w
A T enables us to pose questions about T in terms of linear systems of equations, about which
we know a good deal.
n
m
m
First, T : R → R is one-to-one exactly when T (X) =< 0,0,··· ,0 > in R implies that
n
X =< 0,0,··· ,0 > in R . This is equivalent to asserting that the m × n system
A T X = O
has only the trivial solution X = O. This occurs if and only if n − rank(A T ) = 0, which in turn
occurs if and only if the n columns of A T are linearly independent, since the rank of A T is the
dimension of its row space. This establishes the following.
THEOREM 7.19

n
m
Let T : R → R be a linear transformation. Then the following conditions are equivalent:
1. T is one-to-one.
2. rank(A T ) = n.
3. The columns of A T are linearly independent.

This can be checked for T Example 7.36, with A T given in Example 7.38. There A T was a
7 × 4 matrix having rank 4, and T was one-to-one.
m
A T will also tell us if T is onto. For T to be onto, for each B in R , there must be some X
n
in R such that T (X) = B. This means that the m × n system A T X = B must have a solution for
m
each B, and this is equivalent to the columns of A T forming a spanning set for R . We therefore
have the following.

THEOREM 7.20
n
m
Let T : R → R . Then the following are equivalent.
1. T is onto.
m
2. The system A T (X) = B has a solution for each B in R .
m
3. The columns of A T span R .
.
.
4. rank(A T ) = rank([A T .B] for each B in R .
m

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October 14, 2010 14:23 THM/NEIL Page-245 27410_07_ch07_p187-246